Two capillary tubes of the same diameter are kept vertically in two different liquids whose densities are in…

Two capillary tubes of the same diameter are kept vertically in two different liquids whose densities are in the ratio $4: 3$. The rise of liquid in two capillaries is ' $h_1$ ' and ' $h h_2$ ' respectively. If the surface tensions of liquids are in the ratio $6: 5$, the ratio of heights $\left(\frac{h_1}{h_2}\right)$ is (Assume that their angles of contact are same)
  1. $0.4$
  2. $0.5$
  3. $0.8$
  4. $0.9$

Solution

Given: Density ratio: $\frac{\rho_1}{\rho_2}=\frac{4}{3}$ and surface tension ratio: $\frac{\mathrm{T}_1}{\mathrm{~T}_2}=\frac{6}{5}$ $\therefore \quad$ Rise of liquid in a capillary: $\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\mathrm{r} \rho \mathrm{g}}$ For constant $\theta, \mathrm{r}$ and $\mathrm{g}$, $\begin{aligned} \mathrm{h} & \propto \frac{\mathrm{T}}{\rho} \\ \therefore \quad & \frac{\mathrm{h}_1}{\mathrm{~h}_2}=\frac{\mathrm{T}_1 \rho_2}{\mathrm{~T}_2 \rho_1}=\frac{6 \times 3}{5 \times 4}=0.9 \end{aligned}$

Asked in: MHT CET 2023 (11 May Shift 1)

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